Q 12-03-190JEE MainJEE Main 2021 (31 Aug, Shift 1)Medium
The voltage drop across $15\ \Omega$ resistance in the given figure will be ______ V.
Numerical value type. Enter your answer.
Answer: 6
Upper branch: $(4\parallel4) + 2 + (15\parallel10) = 2 + 2 + 6 = 10\ \Omega$.
Lower branch: $(8\parallel8) + (12\parallel12) = 4 + 6 = 10\ \Omega$.
The two branches in parallel: $5\ \Omega$. Total with $1\ \Omega$: $6\ \Omega$, so $I = \dfrac{12}{6} = 2$ A, and each branch carries $1$ A.
Voltage across the $15\parallel10$ combination $= 1\times6 = 6$ V.
Solution by Sreeraj P, M.Sc Physics